Question

given sin θ = $\frac{5}{6}$, what is cos θ? cos θ = $\frac{sqrt{?}}{}$ report your answer in simplest form

Explanation:

Step1: Recall Pythagorean identity

$\sin^{2}\theta+\cos^{2}\theta = 1$

Step2: Substitute $\sin\theta$ value

$(\frac{5}{6})^{2}+\cos^{2}\theta=1$, so $\frac{25}{36}+\cos^{2}\theta = 1$.

Step3: Solve for $\cos^{2}\theta$

$\cos^{2}\theta=1 - \frac{25}{36}=\frac{36 - 25}{36}=\frac{11}{36}$.

Step4: Find $\cos\theta$

$\cos\theta=\pm\frac{\sqrt{11}}{6}$. Since the triangle in the picture seems to be in the first - quadrant (where cosine is positive for the acute angle $\theta$), $\cos\theta=\frac{\sqrt{11}}{6}$.

Answer:

$\frac{\sqrt{11}}{6}$